Problem: $ \left(\dfrac{1}{64}\right)^{-\frac{2}{3}}$
Solution: $= 64^{\frac{2}{3}}$ $= \left(64^{\frac{1}{3}}\right)^{2}$ To simplify $64^{\frac{1}{3}}$ , figure out what goes in the blank: $\left(? \right)^{3}=64$ To simplify $64^{\frac{1}{3}}$ , figure out what goes in the blank: $\left({4}\right)^{3}=64$ so $ 64^{\frac{1}{3}}=4$ So $64^{\frac{2}{3}}=\left(64^{\frac{1}{3}}\right)^{2}=4^{2}$ $= 4\cdot4$ $= 16$